Answer:
Elastic potential energy, E = 200 J
Explanation:
It is given that,
Spring constant, K = 4 N/m
initial stretching in the spring, x = 5 m
Finally, it is stretched an additional 5 m i.e. x' = 5 m
Let E is the elastic energy in the spring after Varg stretches the spring. it is given by :


E = 200 J
So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.
Answer:
I would say the answer is the wave of 21.000Hz
Explanation:
Because it has more frequency, and as more frequency you add, the time or longer period also increases.
KE = 1/2 x 80 x 60^2
KE = 144000
There’s 10mm in a cm: 22mm
See projectiles are very simple unless you understand its core concepts....projectile is nothing just mixture of upward motion and horizontal motion....
THE KEY IS FORGET THE NAME PROJECTILE...ITS JUST HORIZONTAL MOTION + VERTICAL MOTION